Author Topic: VCAA 2010 Exam 1 (Read 670 times) Tweet Share

captainamy · « **on:** November 06, 2012, 05:37:32 pm »

Hey guys, for question 9b of this exam, I tried integration with recognition but didn't seem to get the correct answer and the solutions just confused me even more. Help please?

Laecs · « **Reply #1 on:** November 06, 2012, 05:54:59 pm »

Hey so, the graph they have drawn is for $x{\log e^x$

Upon solving for the derivative through the product rule and getting:

$x+2x\log e^x$

You're correct in approaching it with integration by recognition, i assume your mistake is assuming the graph is for $x^2\log e^x$ and not $x{\log e^x$

So by integration by recognition we start with:

$\int^3_1 2x\log e^x + x \,dx = \left[x^2\log e ^x \right]_1^3$

We can then take the 2 out, and take the antiderivative of the $x$ , as we are trying to solve for $\int^3_1 x\log e^x \,dx$

Thus giving $2\int^3_1 x\log e^x \,dx$ + $\left[\dfrac{x^2}{2} \right]_1^3=\left[x^2\log e ^x \right]_1^3$

From here it should be relatively simple to just..

$\int^3_1 x\log e^x \,dx$ = $\dfrac{1}{2}\left([x^2\log e^x]_1^3) - [\dfrac{x^2}{2}]_1^3\right)$

$\dfrac{9}{2}\log e^3-2$

I hope this helped.

BubbleWrapMan · « **Reply #2 on:** November 06, 2012, 06:03:36 pm »

Beaten but I typed it up so meh

We want $\int_{1}^{3}x\ln(x)\mathrm{d}x$

We know $\frac{\mathrm{d} }{\mathrm{d} x}(x^2\ln(x))=2x\ln(x)+x$

Integrating both sides from x = 1 to 3 gives

$\int_{1}^{3}\frac{\mathrm{d} }{\mathrm{d} x}(x^2\ln(x))\mathrm{d}x=\int_{1}^{3}(2x\ln(x)+x)\mathrm{d}x$

$\Rightarrow \left [ x^2\ln(x) \right ]_{1}^{3}=\int_{1}^{3}2x\ln(x)\mathrm{d}x+\int_{1}^{3}x\mathrm{d}x$ (since $x^2\ln(x)$ is an antiderivative of $\frac{\mathrm{d} }{\mathrm{d} x}(x^2\ln(x))$ )

$\Rightarrow \left [ x^2\ln(x) \right ]_{1}^{3}=2\int_{1}^{3}x\ln(x)\mathrm{d}x+\left [ \frac{1}{2}x^{2} \right ]_{1}^{3}$

$\Rightarrow \int_{1}^{3}x\ln(x)\mathrm{d}x=\frac{1}{2}\left (\left [ x^2\ln(x) \right ]_{1}^{3}-\left [ \frac{1}{2}x^{2} \right ]_{1}^{3} \right )$

This is basically recognition, but it does the process of finding the antiderivative and the definite integral in one step. You can find the antiderivative through recognition and then find the definite integral as well. How did you do it, out of curiosity?

captainamy · « **Reply #3 on:** November 06, 2012, 06:07:50 pm »

Ahh, yeah I didn't notice that the derivative worked out in 9a was for x^2 ln (x). Thanks a lot ! And don't worry I still understood everything all the same, I appreciate the help !

captainamy · « **Reply #4 on:** November 06, 2012, 06:12:53 pm »

Oh thanks ClimbTooHigh, and I approached the question much like Laecs but I initially thought the derivative found in 9a was for x ln(x) instead.

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